Question

determine whether triangles abc and dbe are similar by following these steps.

1. to use sas similarity, ab and cb must be proportional to db and eb, respectively.
2. find and simplify this ratio: ab/db = 3
3. find and simplify this ratio: cb/eb = 4
4. are the corresponding side lengths proportional?

no
yes
check

Explanation:

Step1: Find AB and DB lengths

From the diagram, \( AB = AD + DB = 10.4 + 5.2 = 15.6 \), \( DB = 5.2 \).

Step2: Calculate \( \frac{AB}{DB} \)

\( \frac{AB}{DB} = \frac{15.6}{5.2} = 3 \).

Step3: Find CB and EB lengths

\( CB = CE + EB = 23.4 + 7.8 = 31.2 \), \( EB = 7.8 \).

Step4: Calculate \( \frac{CB}{EB} \)

\( \frac{CB}{EB} = \frac{31.2}{7.8} = 4 \). Wait, no, wait, let's re - check. Wait, maybe I misread the diagram. Wait, actually, looking at the diagram again, \( AB = AD + DB \), \( AD = 10.4 \), \( DB = 5.2 \), so \( AB = 10.4 + 5.2=15.6 \), \( DB = 5.2 \), so \( \frac{AB}{DB}=\frac{15.6}{5.2} = 3 \). Then \( CB=CE + EB \), \( CE = 23.4 \), \( EB = 7.8 \), so \( CB=23.4 + 7.8 = 31.2 \), \( EB = 7.8 \), \( \frac{CB}{EB}=\frac{31.2}{7.8}=4 \)? Wait, no, that can't be. Wait, maybe the diagram is such that \( AB = AD + DB \), but maybe \( AD = 10.4 \), \( DB = 5.2 \), so \( AB=10.4 + 5.2 = 15.6 \), \( DB = 5.2 \), ratio \( 15.6/5.2 = 3 \). Then \( CB=CE + EB \), \( CE = 23.4 \), \( EB = 7.8 \), so \( CB = 23.4+7.8 = 31.2 \), \( EB = 7.8 \), \( 31.2/7.8 = 4 \)? But that would mean the ratios are not equal. Wait, no, maybe I made a mistake. Wait, let's check again. Wait, maybe \( AB = 10.4 + 5.2=15.6 \), \( DB = 5.2 \), so \( 15.6\div5.2 = 3 \). Then \( CB=23.4 + 7.8 = 31.2 \), \( EB = 7.8 \), \( 31.2\div7.8 = 4 \). But that would mean the ratios are different. But wait, maybe the problem has a typo or I misread the diagram. Wait, no, maybe the lengths are \( AB = 10.4+5.2 = 15.6 \), \( DB = 5.2 \), ratio \( 3 \). \( CB=23.4 + 7.8=31.2 \), \( EB = 7.8 \), ratio \( 4 \). But that would mean the corresponding sides are not proportional. But wait, maybe I messed up the sides. Wait, the triangles are ABC and DBE. So angle at B is common. So for SAS similarity, we need \( \frac{AB}{DB}=\frac{CB}{EB} \) and \( \angle B\) common. Let's recalculate:

\( AB = AD + DB=10.4 + 5.2 = 15.6 \), \( DB = 5.2 \), so \( \frac{AB}{DB}=\frac{15.6}{5.2}=3 \).

\( CB=CE + EB = 23.4+7.8 = 31.2 \), \( EB = 7.8 \), so \( \frac{CB}{EB}=\frac{31.2}{7.8}=4 \). Wait, that's a problem. But maybe the diagram is different. Wait, maybe \( AB = 10.4 \), \( DB = 5.2 \), so \( \frac{AB}{DB}=\frac{10.4}{5.2}=2 \)? No, the problem says \( AB/DB = 3 \) in step 3. Wait, maybe the length of \( AB \) is \( 15.6 \) (10.4 + 5.2) and \( DB = 5.2 \), so 15.6/5.2 = 3. Then \( CB \) is \( 23.4+7.8 = 31.2 \), \( EB = 7.8 \), 31.2/7.8 = 4. But 3≠4, so the answer should be no. But wait, maybe I made a mistake. Wait, let's check the numbers again. 15.6 divided by 5.2: 5.23 = 15.6, correct. 31.2 divided by 7.8: 7.84 = 31.2, correct. So the ratios are 3 and 4, which are not equal. So the corresponding side lengths are not proportional.

Answer:

no